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31 Jan 2022, 03:15
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Difficulty:
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Question Stats:
92% (01:00) correct
8%
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In the figure above, ABCD is a square. If AC is 8 units long, what is the perimeter of the square?
(A) \(16\)
(B) \(12 \sqrt{2}\)
(C) \(24\)
(D) \(16 \sqrt{2}\)
(E) \(30\)
Attachment:
1.jpg [ 6.82 KiB | Viewed 2392 times ]
31 Jan 2022, 08:03
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AC is the diagonal of the square.
diagonal square =x sqrt(2)= 8 (where x is the side)
==> x = 8/sqrt(2)
Perimeter = 4x = (4 * 8) / sqrt(2)
Now we want to get rid of the sqrt(2)
i) Break down 4 into 2 * 2.
ii) Then break down one of the 2's into sqrt(2) * sqrt(2)
iii) Cancel out one of the sqrt(2)'s with the sqrt(2) from the denominator
This leaves us with:
2 * sqrt(2) * 8 = 16 sqrt(2)
===> D
diagonal square =x sqrt(2)= 8 (where x is the side)
==> x = 8/sqrt(2)
Perimeter = 4x = (4 * 8) / sqrt(2)
Now we want to get rid of the sqrt(2)
i) Break down 4 into 2 * 2.
ii) Then break down one of the 2's into sqrt(2) * sqrt(2)
iii) Cancel out one of the sqrt(2)'s with the sqrt(2) from the denominator
This leaves us with:
2 * sqrt(2) * 8 = 16 sqrt(2)
===> D
31 Jan 2022, 08:10
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Let AB = BC = CD = AD = a.
\(a^{2}\) +\( a^{2}\) = \(AC^{2}\)
2\(a^{2}\) = 64
a = 4\(\sqrt{2}\)
P = 4a = 16\(\sqrt{2}\)
Option D
\(a^{2}\) +\( a^{2}\) = \(AC^{2}\)
2\(a^{2}\) = 64
a = 4\(\sqrt{2}\)
P = 4a = 16\(\sqrt{2}\)
Option D
